3.199 \(\int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=78 \[ \frac {\sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

1/2*sin(d*x+c)/b^2/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+1/2*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/b^2/d/(b*c
os(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {18, 3768, 3770} \[ \frac {\sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 b^2 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(5/2)),x]

[Out]

(ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(2*b^2*d*Sqrt[b*Cos[c + d*x]]) + Sin[c + d*x]/(2*b^2*d*Cos[c + d*x]
^(3/2)*Sqrt[b*Cos[c + d*x]])

Rule 18

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m - 1/2)*b^(n + 1/2)*Sqrt[a*v])/Sqrt[b*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \sec ^3(c+d x) \, dx}{b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {\sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \int \sec (c+d x) \, dx}{2 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {\tanh ^{-1}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 55, normalized size = 0.71 \[ \frac {\sqrt {b \cos (c+d x)} \left (\sin (c+d x)+\cos ^2(c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{2 b^3 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(5/2)),x]

[Out]

(Sqrt[b*Cos[c + d*x]]*(ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + Sin[c + d*x]))/(2*b^3*d*Cos[c + d*x]^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 207, normalized size = 2.65 \[ \left [\frac {\sqrt {b} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b^{3} d \cos \left (d x + c\right )^{3}}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(b)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*
x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*co
s(d*x + c)^3), -1/2*(sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*
x + c)^3 - sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^3)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*cos(d*x + c))^(5/2)*sqrt(cos(d*x + c))), x)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 104, normalized size = 1.33 \[ -\frac {\left (\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-\sin \left (d x +c \right )-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-\sin \left (d x +c \right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{2 d \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x)

[Out]

-1/2/d*(cos(d*x+c)^2*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-cos(d*x+c)^2*ln(-(-sin(d*x+c)-1+cos(d*x+c))/si
n(d*x+c))-sin(d*x+c))*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2)

________________________________________________________________________________________

maxima [B]  time = 1.56, size = 688, normalized size = 8.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/4*(4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(
4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (2*(2*cos(2*d*x + 2*
c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)
*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2
*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) +
1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c)))^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x +
2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))/((b^2*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + b
^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x +
2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c))*sqrt(b)*d)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(5/2)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)**(1/2)/(b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________